$\begingroup$

Is there any way to simplify this function? Or is this the simplest equation? : AB + A'+B'

$\endgroup$ 7

5 Answers

$\begingroup$

$$AB+A'+B' = AB+ \overline{AB} = 1$$

$\endgroup$ $\begingroup$

$$0\cdot0+1+1=1\\1\cdot0+0+1=1\\0\cdot1+1+0=1\\1\cdot1+0+0=1$$

$\endgroup$ $\begingroup$

A general technique in boolean algebra is to add or multiply some term by "$P+P$'" to make it similar to other terms so that they can be simplified. In this case:

$AB+A'(B+B')+B'$

$=(A+A')B+A'B'+B'$

$=B+A'B'+(A+A')B'$

$=B+A'B'+AB'+A'B'$

$=B+A'B'+AB'$

$=B+(A+A')B'$

$=B+B'$.

But in complicated cases it may not simplify immediately.

Note that this solution uses only identity, inverse, idempotency and distributivity and so avoids using De Morgan's laws unlike the other solutions. De Morgan's laws, which are extremely useful of course, can be proven in a similar way if desired. Alternatively one can use truth tables to verify all basic rules.

$\endgroup$ $\begingroup$

F = AB + A'+B'
F' = (A'+B').AB = A'AB + ABB' = 0

Again if I take F = F'' = 1 , 1 will the the result.

$\endgroup$ 4 $\begingroup$

AB + A' + B'

Expand missing terms.

AB + A' (B' + B) + B' (A' + A) - Complement Law

AB + A'B' + A'B + A'B' + AB'

Eliminate duplicates and rearrange.

A'B' + A'B + AB' + AB

A' (B' + B) + A (B' + B) - Complement Law

A' + A

1

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy